In the tetrahedron $ABCD$, perpendiculars $AB'$, $AC'$, $AD'$ are dropped from vertex $A$, on the plane dividing the dihedral angles at the edges $CD$, $BD$, $BC$ in half. Prove that the plane $(B'C'D' )$ is parallel to the plane $(BCD)$.
Source: - All-Russian MO 2006 Regional (R4) 11.6
Tags: geometry, 3D geometry, tetrahedron
In the tetrahedron $ABCD$, perpendiculars $AB'$, $AC'$, $AD'$ are dropped from vertex $A$, on the plane dividing the dihedral angles at the edges $CD$, $BD$, $BC$ in half. Prove that the plane $(B'C'D' )$ is parallel to the plane $(BCD)$.